The Story of Louis Pósa

Ross Honsberger

WOULD LIKE TO TELL YOU something of the life and works
of a remarkable young Hungarian named Louis Pósa (pronounced pO.sha),
who was born in the late 1940's. When quite young he attracted the attention
of the eminent Hungarian mathematician Paul Erdös (pronounced air.dish),
who did much to help him develop. Erdös has recently written and spoken
about some of the child prodigies he has known and I would like to tell
you his story of Pósa.

In case you are not familiar with
Erdös, let me first tell you a little about him. He is now about 60 years
old and he has been for many years an internationally known mathematician.
His three great interests are combinatorics, number theory, and geometry.
He is a problem solver rather than a theory builder, although a fair number
of his more than 500 mathematical papers exceed 100 pages in length. For decades
he has travelled the world's universities, seldom visiting anywhere for more
than a few months. He spent the Fall term of 1970 at the University of Waterloo,
and during this visit he told us about the Hungarian prodigies. Except for
a few minor changes and additions, the following is his story of Pósa.

Erdös' Story. "I will talk
about Pósa who is now 22 years old and the author of about 8 papers.
I met him before he was 12 years oild. When I returned from the United States
in the summer of 1959 I was told about a little boy whose mother was a mathematician
and who knew quite a bit about high school mathematics. I was very interested
and the next day I had lunch with him. While Pósa was eating his soup
I asked him the following question: Prove that if you have n+1 positive integers
less than or equal to 2n, some pair of them are relatively prime. It is quite
easy to see that the claim is not true of just n such numbers, because no
two of the n even numbers up to 2n are relatively prime.; Actually I discovered
this simple result some years ago but it took me about ten minutes to find
the really simple proof. Pósa sat there eating his soup and then after
half a minute or so he said "If you have n+1 positive intergers less than
or equal to 2n, some two of them will have to be consecutive and thus relatively
prime." Needless to say, I was very much impressed, and I venture to class
this on the same level as Gauss' summation of the positive integers up to
100 when he was just 7 years old."

At this point Erdös discussed
a few problems in graph theory which he gave to Pósa. In order to avoid
any misunderstandings, let us interject here a brief introduction to this
material.

By a graph we do not mean anything
connected with axes and coordinates. A graph consists of a collection
of vertices (points) and a set of edges, each joining a pair of vertices.
How the graph is pictured on paper is not essential. The edges may be drawn
as straight lines or curves, and it is immaterial whether they are drawn so
as to intersect or not. Points of intersection obtained by edges which cross
do not count as vertices. Only the given vertices are the vertices of the
graph. Also a graph need not contain every possible edge which could be drawn,
that is, there are in general many different graphs with the same set of vertices.

A loop is an edge both of whose
endpoints are the same vertex. Multiple edges occur when two or more edges
join the same pair of vertices. In a general graph both loops and multiple
edges are permitted. In a simple (or strict) graph neither loops nor multiple
edges may occur. Throughout Erdös' story and the remainder of this essay
the unmodified word graph is intended to mean simple graph. We continue now
with the story.

"From that time onward I worked
systematically with Pósa. I wrote to him of problems many times during
my travels. While still 11 he proved the following theorem which I proposed
to him: A graph with 2n vertices and n2 + 1 edges must contain
a triangle. Actually this is a special case of a well-known theorem of Turán,
which he worked out in 1940 in a Hungarian labor camp. I also gave him the
following problem: Consider an infinite series whose nth term is the fraction
with numerator 1 and denominator the lowest common multiple of the integers
1, 2, ...., n; prove that the sum is an irrational number. This is not very
difficult, but it is certainly surprising that a 12 year old child could do
it.

"When he was just 13, I explained
to him Ramsey's theorem for the case k=2: Suppose you have a graph with an
infinite number of vertices; there is either an infinite set of vertices every
two of which are joined by an edge, or there is an infinite set of vertices
no two of which are joined by an edge. (Incidentally, this theorem is the
discovery of the late Frank Ramsey, a brother of the present Archbishop of
Canterbury.) It took about 15 minutes for Pósa to understand it. Then
he went home, thought about it all evening, and before going to sleep he had
found a proof. "By the time Pósa was about 14 years old you could talk
to him as a grown-up mathematician. It is interesting to note that he had
some difficulty with calculus. He never liked geometry, and he never wanted
to bother with anything that did not really excite him. At anything that did
interest him, however, he was extremely good. Our first joint paper was written
when he was 14 1/2. Pósa wrote several significant papers by himself,
some of which still have a great deal of effect. His best known paper, on
Hamiltonian circuits, for which he received international acclaim, he wrote
when he was only 15!

"The first theorem that he discovered
and proved by himself which was new mathematics was the following: A graph
with n vertices (n = 4) and 2n - 3 edges must contain a circuit
with a diagonal. This result is the best possible, because for every n,
one can construct a graph with n vertices and 2n - 4 edges which fails to
have a circuit with a diagonal.

"A problem which I had previously
solved is the following: A graph with n vertices (n =6) and 3n-5 edges
must contain two circuits which have no vertices in common. I told Pósa
of the problem, and in a few days he had a very simple proof which was miles
ahead of the complicated one I had come up with. A most remarkable thing for
a child of 14.

"Pósa also found a beautiful
proof that every graph with n vertices and n+ 4 edges contains two circuits
which have no edges in common. (This also holds for general graphs.)

"I would like to make a few conjectures
why there are so many child prodigies
in Hungary. First of all there has been for at least 80 years a mathematical
periodical for high-school students. Then there are many mathematical competitions.
The Eötvöse-Kurshák competition goes back 75 years. After
the first world war a new competition was begun for students just completing
high school, and after the second war several new ones were started.

"A few years ago a different kind
of competition was started. It is held on television. Bright high school students
compete in doing questions in a given period of time. The questions are usually
very ingenious and the solutions are judged by a panel of leading mathematicians
such as Alexits, Turån, and Hajos. It seems many people watch these
competitions with great interest even though they do not understand the problems.

"In Hungary a few years ago a
special high school, the Michael Fazekas High School, was opened in Budapest
for children who are gifted in mathematics. The school started just when Pósa
was due to go to high school. He liked the school very much, so much so, in
fact, that he refused to leave it for entrance into university two years early.
Soon after attending Fazekas High School, Pósa was telling me of other
boys in his class who he thought were better at elementary mathematics than
he was. Among these boys was the now prominent Lovász."

It seems appropriate to complement
the story of Pósa with a sample of his work. Accordingly, let us work
through his beautiful proof of Dirac's theorem on Hamiltonian circuits.

Dirac's Theorem

Graphs have many interesting properties.
In 1857 the Irish genius William Rowan Hamilton invented a game of travelling
around the edges of a graph from vertex to vertex. Given a particular graph
to begin with, the object of the game was to find a path in the graph which
passes through each vertex exactly once.

Of course, in some graphs such
a path does not exist. If, in addition to finding such a path, one can arrange
to make the last vertex the same as the first, he obtains a Hamiltonian circuit,
not just a Hamiltonian path. While a Hamiltonian circuit always provides a
Hamiltonian path, upon the deletion of any edge, a Hamiltonian path may not
lead to a Hamiltonian circuit (it depends on whether or not the first and
last vertices of the path happen to be joined by an edge in the graph).

There are some necessary and some
sufficient conditions known for the existence of a Hamiltonian circuit in
a graph. For example, it is necessary to have as many edges as vertices, and
it is sufficient to have all possible edges actually as vertices, and it is
sufficient to have all possible edges actually in the graph. However, even
in 1972 there are no known necessary and sufficient conditions.

In 1952 the European mathematician
G.A. Dirac came up with the following simple sufficient condition. (The "degree"
of a vertex is simply the number of edges that occur at the vertex.)

DIRAC's THEOREM. A graph with
n vertices (n Ñ 3) in which each vertex has degree at least n/2 has a Hamiltonian
circuit.

In 1962 Pósa produced the
following proof. He proceeds indirectly. Suppose the given graph G which has
n vertices and every degree at least n/2 is non-Hamiltonian, that is, does
not possess a Hamiltonian circuit. We now proceed to a contradiction.

Firstly we embed G in a saturated
non-Hamiltonian graph G' as follows:

Clearly not all possible edges
occur in G or else it would be Hamiltonian already. Consider, then, the insertion
of an edge which is missing from G.

If this insertion does not complete
a Hamiltonian circuit then leave the edge in the graph. If it does complete
a Hamiltonian circuit, take the edge back out. Go around G doing this at each
missing edge. The order in which one tries the missing edges is unimportant.
At the end of this tour, the resulting graph G' will still be non-Hamiltonian
since no Hamiltonian circuit is ever allowed to remain intact, and yet it
will be saturated in the sense that the addition now of any missing edge will
be have to complete a Hamiltonian circuit (otherwise we would have left the
edge in the graph when it was tried during our rounds).

Adding edges certainly does not
diminish the degree of any vertex. Hence every vertex of G' has degree at
least n/2.

Since G' is non-Hamiltonian, it
cannot contain all possible edges. Suppose the edge between vertices v1
and vn is missing. Putting in this edge would complete a Hamiltonian
circuit because G' is saturated. Thus, even without the edge v1vn,
the graph G' must contain a Hamiltonian path from v1 to vn
(inserting the edge v1vn merely completes this path
into a circuit). Let us denote the order of the n vertices in this path by
v1, v2, v3, ..., vn-1, vn.

Now if the vertex v1
happens to be joined to the vertex vi, we ask whether the vertex
vn could possibly be joined to vi-1? The answer is "no."
For, if vn were joined to vi-1, then G' would contain
the Hamiltonian circuit

v1vivi+1
... vnvi-1vi-2 ... v2v1.

This is impossible since G' is
non-Hamiltonian.

While v1
is not joined to vn, we do know that v1 has degree at
least n/2. Thus v1 is joined to at least n/2 vertices vi
where i runs from 2 to n - 1. Consequently, there are at least n/2 vertices
vi-1, where i - 1 ranges from 1 to n - 2, to which vn cannot possibly be joined.
Since loops are not permitted, vn cannot be joined to itself, either.
Altogether, then, there are more than n/2 vertices to which vn
cannot be joined. This leaves fewer than n/2 vertices to which vn
can be joined, contradicting the given fact that its degree is at least n/2.
(QED)

We conclude with the statement
of four recent theorems in the field of Hamiltonian circuits. The final one
is the work of V. Chvátal, an energetic young Czech mathematician,
who produced it during his stay at the University of Waterloo, where shortly
before he had earned his Doctorate Degree.

DIRAC'S THEOREM (1952). A graph
with n vertices (n=3) in which each vertex has degree at least n/2 has
a Hamiltonian circuit.

ORE'S THEOREM (1960). If n=3
and, for every pair of vertices that are not joined by an edge, the sum of
the degrees is at least n, then the graph is Hamiltonian.

PÓSA'S THEOREM (1962).
Let G be a simple graph with n vertices. If for every k in 1= k 
(n - 1)/2, the number of vertices of degree not exceeding k is less than k,
and if for n odd the number of vertices with degree not exceeding (n - 1)/2
does not exceed (n-1)/2, then G contains a Hamiltonian circuit.

CHAVTÁL'S THEOREM (1970).
Let the graph G have vertices with degrees d1, d2, ...,
dn, written in non-decreasing order. If for every i<n/2 we have
either di = i + 1 or dn-1 = n-i, then the graph is Hamiltonian.

Exercises

1. Find a Hamiltonian circuit
in each of the following graphs:

2. Prove that there is no Hamiltonian
circuit in each of the following graphs:

3. Prove there is no Hamiltonian
path in either of the graphs:

4. Which of the following
graphs have Hamiltonian circuits, and which have only Hamiltonian paths?

5. Six points in general position
are given in three-dimensional space (that is, no 3 are collinear and no 4
are coplanar). The C (6,2) = 15 segments joining them in pairs are colored
individually either red or blue at random. Prove that some triangle has all
its sides the same color.

6. A set of moves in chess which
takes a knight successively through all 64 squares is called a knight's tour.
If the knight can go from the last square to the first one in one move, and
thus go all around again, the tour is called re-entrant. A re-entrant knight's
tour corresponds to a Hamiltonian circuit in a graph which has a vertex for
each square and an edge joining the vertices representing squares X and Y
if and only if a knight can go from X to Y in one move. Show that there is
no re-entrant knight's tour on any chessboard which has dimensions 4 by n,
n a natural number.

8. Try your hand at proving some
of the theorems Pósa worked through as a teenager. For example, prove
that a graph with n vertices and n+4 edges contains two circuits with no edge
in common. The proof of this is contained in the joint paper of Erdös
and Pósa "On the Maximum Number of Disjoint Circuits of a Graph", Publications
Mathematicae, Debrecen, 1962, 3-12, especially page 9. This journal may be
listed under Kossuth Lajos Tudomanyegyetem, Mathematika Intezet.

9. Prove Ramsey's Theorem. If
you have trouble, consult the following plan. Consider an infinite sequence
of vertices, 1, 2, 3, ...

Suppose each pair is joined by
an edge. The edge joining vertices n and n + p, p > 0, is considered to
begin at n and end at n + p. Now all edges are colored either red or blue
so that, for every vertex n, all the edges which begin at n have the same
color. If all are red, then n is called a red vertex, etc. Which vertices
are red and which blue is immaterial.

(a) Show in such a sequence there
exists either an infinite set of vertices every two of which are joined by
a red edge, or there exists an infinite set of vertices every two of which
are joined by a blue edge. (Hint: Consider the set of red vertices and the
set of blue vertices; at least one of these sets must be infinite.)

(b) Now complete the proof of
Ramsey's theorem by showing that in every graph which has an infinite number
of vertices one can determine a sequence of vertices as described above.

(Hint: Color all edges of the
graph red. Then insert all the missing edges and color them blue. Take any
vertex at all as vertex 1 of the sequence. Separate into classes R and B the
vertices to which 1 is joined, respectively, by red and blue edges. At least
one of R and B must contain an infinity of vertices. If both do, choose either.
For definiteness, suppose R is infinite. Then take any vertex of R as vertex
2 of the sequence. Repeat
the separation process with respect to vertex 2 and its infinite set B (forget
all about set B). Choose any vertex of an infinite set thus obtained as vertex
3 of the sequence, and so on. From
this sequence Ramsey's theorem is easily deduced by part (a). At the end rub
out all the blue edges.)

10. Let A0, A1, A2, ..., A2n-1
denote, in cyclic order, the vertices of a regular
2n-gon. Let all the sides and diagonals be drawn to give graph G. Prove that
every Hamiltonian circuit of G must contain two edges which are parallel lines
in the diagram.